Question

At high temperatures, sulfur combines with iron to form brown-black iron(II) sulfide:

Fe(s) + SD-Fes(s)
In an experiment, 5.00 g of iron reacts with excess sulfur to produce 6.29 g of iron(II) sulfide. Calculate the percent yield if the maximum
amount of iron(II) sulfide that can be produced is 7.87 g.
79.796
79.9%
100%
63.5%
125%

Answer 1

The percent yield : 79.9%

Further eplanation

Percent yield is the comparison of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

Fe(s)+S(s)⇒FeS(s)

The reaction produces 6.29 g of iron(II) sulfide⇒an actual yield

The maximum amount that can be produced is 7.87 g ⇒ A theoretical yield

`\tt \%yield=(actual)/(theoretical)* 100\%\\\\\%yield=(6.29)/(7.87)* 100\5=79.9\%`